Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.
Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
Consider this example:
= = = = - = Cows facing right --> = = = = - = = = = = = = = = 1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
6 10 3 7 4 12 2
5【题意】n头奶牛站成一排，从左往右看，只要右边高度<=自己高度的都可以看到，但是一旦遇到比自己高的，则它后面的奶牛都看不到，即计数停止，要计算所有牛向右所能看到的数目和。（奶牛哪知道哪边是左哪边是右~~~汗，，，）
【AC代码】
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=8e4+5;int main()
{int m,n;ll a[N],vis[N];while(~scanf("%d",&m)){ll ans=0; //long long 可以通过memset(a,0,sizeof(a));memset(vis,0,sizeof(vis));for(int i=1;i<=m;i++){scanf("%lld",&a[i]);vis[i]=i;}for(int i=m-1;i>0;i--){int s=i;while(a[s+1]<a[i]&&s<m)//相等也看不到s=vis[s+1];vis[i]=s;}for(int i=1;i<=m;i++)ans+=(vis[i]-i);printf("%lld\n",ans);}return 0;
}